[英]selected value get from db into dropdown select box option using php mysql error
我需要从 db 中获取选定的值到选择框中。 请告诉我怎么做。 这是代码。 注意:'options' 值取决于类别。
<?php
$sql = "select * from mine where username = '$user' ";
$res = mysql_query($sql);
while($list = mysql_fetch_assoc($res)){
$category = $list['category'];
$username = $list['username'];
$options = $list['options'];
?>
<input type="text" name="category" value="<?php echo '$category' ?>" readonly="readonly" />
<select name="course">
<option value="0">Please Select Option</option>
<option value="PHP">PHP</option>
<option value="ASP">ASP</option>
</select>
<?php
}
?>
我认为您正在寻找以下代码更改:
<select name="course">
<option value="0">Please Select Option</option>
<option value="PHP" <?php if($options=="PHP") echo 'selected="selected"'; ?> >PHP</option>
<option value="ASP" <?php if($options=="ASP") echo 'selected="selected"'; ?> >ASP</option>
</select>
我能想到的最简单的方法如下:
<?php
$selection = array('PHP', 'ASP');
echo '<select>
<option value="0">Please Select Option</option>';
foreach ($selection as $selection) {
$selected = ($options == $selection) ? "selected" : "";
echo '<option '.$selected.' value="'.$selection.'">'.$selection.'</option>';
}
echo '</select>';
代码基本上将所有选项放在一个数组中,这些选项在 foreach 循环中被调用。 循环检查您的 $options 变量是否与它所在的当前选择匹配,如果匹配,则 $selected 将 = 选择,如果不匹配,则将其设置为空白。 最后返回包含数组中选择的选项标签,如果该特定选择等于您的 $options 变量,则将其设置为所选选项。
例如..下次请使用 mysqli() 因为 mysql() 已被弃用。
<?php
$select="select * from tbl_assign where id='".$_GET['uid']."'";
$q=mysql_query($select) or die($select);
$row=mysql_fetch_array($q);
?>
<select name="sclient" id="sclient" class="reginput"/>
<option value="">Select Client</option>
<?php $s="select * from tbl_new_user where type='client'";
$q=mysql_query($s) or die($s);
while($rw=mysql_fetch_array($q))
{ ?>
<option value="<?php echo $rw['login_name']; ?>"<?php if($row['clientname']==$rw['login_name']) echo 'selected="selected"'; ?>><?php echo $rw['login_name']; ?></option>
<?php } ?>
</select>
从下拉列表中选择值。
<select class="form-control" name="category" id="sel1">
<?php foreach($data as $key =>$value) { ?>
<option value="<?php echo $data[$key]->name; ?>"<?php if($id_name[0]->p_name==$data[$key]->name) echo 'selected="selected"'; ?>><?php echo $data[$key]->name; ?></option>
<?php } ?>
</select>
它会在您的选项中添加额外的内容,但您的问题将得到解决。
<?php
if ($editing == Yes) {
echo "<option value=\".$MyValue.\" SELECTED>".$MyValue."</option>";
}
?>
最好的代码和简单
<select id="example-getting-started" multiple="multiple" name="category">
<?php
$query = "select * from mine";
$results = mysql_query($query);
while ($rows = mysql_fetch_assoc(@$results)){
?>
<option value="<?php echo $rows['category'];?>"><?php echo $rows['category'];?></option>
<?php
}
?>
</select>
你也可以这样做......
<?php $countryname = $all_meta_for_user['country']; ?>
<select id="mycountry" name="country" class="user">
<?php $myrows = $wpdb->get_results( "SELECT * FROM wp_countries order by country_name" );
foreach($myrows as $rows){
if( $countryname == $rows->id ){
echo "<option selected = 'selected' value='".$rows->id."'>".$rows->country_name."</option>";
} else{
echo "<option value='".$rows->id."'>".$rows->country_name."</option>";
}
}
?>
</select>
使用 PDO
<?php
$username = "root";
$password = "";
$db = "db_name";
$dns = "mysql:host=localhost;dbname=$db;charset=utf8mb4";
$conn = new PDO($dns,$username, $password);
$conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
$sql = "select * from mine where username = ? ";
$stmt1 = $conn->prepare($sql);
$stmt1->execute(array($_POST['user']));
$all = $stmt1->fetchAll(); ?>
<div class="controls">
<select data-rel="chosen" name="degree_id" id="selectError">
<?php
foreach($all as $nt) {
echo "<option value =$nt[id]>$nt[name]</option>";
}
?>
</select>
</div>
只需添加一个额外的隐藏选项并从数据库中打印选定的值
<option value="<?php echo $options;?>" hidden><?php echo $options;?></option>
<option value="PHP">PHP</option>
<option value="ASP">ASP</option>
答案很简单。 当您从下拉列表中传递值时。
就像其他一样使用。
例如:
foreach($result as $row) {
$GLOBALS['output'] .='<option value="'.$row["dropdownid"].'"'.
($GLOBALS['passselectedvalueid']==$row["dropwdownid"] ? ' Selected' : '').'
>'.$row['valueetc'].'</option>';
}
<?php
$sql = "select * from mine where username = '$user' ";
$res = mysql_query($sql);
while($list = mysql_fetch_assoc($res)){
$category = $list['category'];
$username = $list['username'];
$options = $list['options'];
?>
<input type="text" name="category" value="<?php echo '$category' ?>" readonly="readonly" />
<select name="course">
<option value="0">Please Select Option</option>
<option value="PHP" <?php echo $options == 'PHP' ? 'selected' : ''; ?> >PHP</option>
<option value="ASP" <?php echo $options == 'ASP' ? 'selected' : ''; ?> >ASP</option>
</select>
<?php
}
?>
我正在使用eval() PHP 函数,如下所示:
我的PHP代码:
$selOps1 = $selOps2 = $selOps3 = '';
eval('$selOps'. $dbRow["DBitem"] . ' = "selected";');
然后在我的选择框中,我像这样使用它:
<select>
<option <?=$selOps1?> value="1">big</option>
<option <?=$selOps2?> value="2">Middle</option>
<option <?=$selOps3?> value="3">Small</option>
</select>
$option = $result['semester'];
<option >Select</option>
<option value="1st" <?php if($option == "1st") echo 'selected = "selected"'; ?>>1st</option>
<option value="2nd" <?php if($option == "2nd") echo 'selected = "selected"'; ?>>2nd</option>
<option value="3rd" <?php if($option == "3rd") echo 'selected = "selected"'; ?>>3rd</option>
<option value="4th" <?php if($option == "4th") echo 'selected = "selected"'; ?>>4th</option>
<option value="5th" <?php if($option == "5th") echo 'selected = "selected"'; ?>>5th</option>
<option value="6th" <?php if($option == "6th") echo 'selected = "selected"'; ?>>6th</option>
<option value="7th" <?php if($option == "7th") echo 'selected = "selected"'; ?>>7th</option>
<option value="8th" <?php if($option == "8th") echo 'selected = "selected"'; ?>>8th</option>
</select>
这可能对你有帮助。
?php
$sql = "select * from mine where username = '$user' ";
$res = mysql_query($sql);
while($list = mysql_fetch_assoc($res))
{
$category = $list['category'];
$username = $list['username'];
$options = $list['options'];
?>
<input type="text" name="category" value="<?php echo '$category' ?>" readonly="readonly" />
<select name="course">
<option value="0">Please Select Option</option>
// Assuming $list['options'] is a coma seperated options string
$arr=explode(",",$list['options']);
<?php foreach ($arr as $value) { ?>
<option value="<?php echo $value; ?>"><?php echo $value; ?></option>
<?php } >
</select>
<?php
}
?>
好的,首先我认为,您需要从数据库中获取一些数据并将其显示在选择框中作为选项,然后您需要将该选择选项值保存到数据库中。
function showAllData(){
$connection = mysqli_connect('localhost', 'root', '', 'beva');
$query = "SELECT * FROM users";
$result = mysqli_query($connection, $query);
if(!$result){
die('Query Failed'. mysqli_error());
}
while($row = mysqli_fetch_assoc($result)){
$id = $row['id'];
echo $id;
echo "<option name='$id'>$id</option>";
}
}
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8">
<title>Document</title>
<link href="https://cdn.jsdelivr.net/npm/bootstrap@5.0.0-beta2/dist/css/bootstrap.min.css" rel="stylesheet" integrity="sha384-BmbxuPwQa2lc/FVzBcNJ7UAyJxM6wuqIj61tLrc4wSX0szH/Ev+nYRRuWlolflfl" crossorigin="anonymous">
<script src="https://cdn.jsdelivr.net/npm/bootstrap@5.0.0-beta2/dist/js/bootstrap.bundle.min.js" integrity="sha384-b5kHyXgcpbZJO/tY9Ul7kGkf1S0CWuKcCD38l8YkeH8z8QjE0GmW1gYU5S9FOnJ0" crossorigin="anonymous"></script>
</head>
<body>
<div class="container mt-3 mb-5">
<div class="card">
<form action="html.php" method="post">
<input type="text" name="username" placeholder="Enter Username">
<input type="password" name="password" placeholder="Enter Password">
<select name="se"> <?php showAllData();?> </select>
<input type="submit" name="submit">
</form>
</div>
</div>
</body>
</html>
在这里,您只需调用该函数。 确保您为选择框输入了一个名称<select name="se">
if(isset($_POST['submit'])){
$connection = mysqli_connect('localhost', 'root', '', 'beva');
$username = $_POST['username'];
$password = $_POST['password'];
$id = $_POST['se'];
echo $id;
$query = "UPDATE users SET username = '$username', password = '$password' WHERE id = $id ";
$result = mysqli_query($connection, $query);
echo $query;
if(!$result){
echo $query;
die("Query Failed" . mysqli_error($connection));
}
}
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